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For the reaction4A(g) + 3B(g) ?2C(g) the following data
Chapter 16, Problem 26P(choose chapter or problem)
For the reaction
\(4\mathrm{A}(g)+3\mathrm{B}(g)\longrightarrow2\mathrm{C}(g)\)
the following data were obtained at constant temperature:
\(\begin{array}{crcc} \text { Experiment } & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \end{array} \\ \hline 1 & 0.100 & 0.100 & 5.00 \\ 2 & 0.300 & 0.100 & 45.0 \\ 3 & 0.100 & 0.200 & 10.0 \\ 4 & 0.300 & 0.200 & 90.0 \end{array}\)
(a) What is the order with respect to each reactant?
(b) Write the rate law.
(c) Calculate k (using the data from experiment 1).
Questions & Answers
QUESTION:
For the reaction
\(4\mathrm{A}(g)+3\mathrm{B}(g)\longrightarrow2\mathrm{C}(g)\)
the following data were obtained at constant temperature:
\(\begin{array}{crcc} \text { Experiment } & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \end{array} \\ \hline 1 & 0.100 & 0.100 & 5.00 \\ 2 & 0.300 & 0.100 & 45.0 \\ 3 & 0.100 & 0.200 & 10.0 \\ 4 & 0.300 & 0.200 & 90.0 \end{array}\)
(a) What is the order with respect to each reactant?
(b) Write the rate law.
(c) Calculate k (using the data from experiment 1).
ANSWER:Step 1 of 4
(a) We have to calculate the order with respect to each reactant
Order of reactant A:-
Let us consider the concentration of [A] is m.
1st we have to identify the reaction experiment in which [A] changes but [B] does not i.e in experiment 1 and 2.
\(\text { Rate expt 1 } / \text { Rate expt 2 }=\left(\left[\mathrm{A}_{\text {expt 1 }}\right] /\left[\mathrm{A}_{\text {expt 2 }}\right]\right)^{\mathrm{m}}\)
\(45 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} / 5 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}=\left(0.300 \mathrm{~mol} \mathrm{~L}^{-1} / 0.100 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{\mathrm{m}}\)
\(\Rightarrow 9.00=(3.00)^{\mathrm{m}}\)
Taking log of both sides,
\(\Rightarrow \log (9.00)=\log (3.00)^{\mathrm{m}}\)
\(\Rightarrow \log (9.00)=\mathrm{mlog}(3.00) \)
\(\Rightarrow \mathrm{m}=2\)